Inverse Trig Functions

Inverse of Function
arc-trig functions are bounded to $[- \frac{π}{2}, \frac{π}{2}]$
- $\arcsin (\sin (\frac{4 π}{5})) = \frac{π}{5}$
  - do $π - θ$
  - If it was $\tan$ , you could do $\pm π$
  - Or you can try $\pm 2 π$
If you have multiple inverse trig functions, you can do the following
- This is also useful for if the inverse function doesn't match the normal function
- Sometimes you will need to draw a triangle

Example

$\begin{aligned} \arctan (\frac{1}{2}) + \arctan (\frac{1}{3}) \\ α & = \arctan (\frac{1}{2}) \\ β & = \arctan (\frac{1}{3}) \\ \tan α & = \frac{1}{2} \\ \tan β & = \frac{1}{3} \\ \tan (α + β) & = \frac{\tan α + \tan β}{1 - \tan α \tan β} \\ = \frac{\frac{1}{2} + \frac{1}{3}}{1 - (\frac{1}{2}) (\frac{1}{3})} \\ = 1 \\ α + β & = \frac{π}{4} \end{aligned}$

For hyperbolic trig functions: just use latin letters, and different identities
To find range:

Example

$\begin{aligned} f (x) & = 2 + e^{- \tan^{- 1} (1 + x^{2})} \\ 1 & \leq 1 + x^{2} < \infty \\ \tan^{- 1} (1) & \leq \tan^{- 1} (1 + x^{2}) < \tan^{- 1} (\infty) \\ \frac{π}{4} & \leq \tan^{- 1} (1 + x^{2}) < \frac{π}{2} \\ - \frac{π}{4} & \geq - \tan^{- 1} (1 + x^{2}) > - \frac{π}{2} \\ e^{- π / 4} & \geq e^{- \tan^{- 1} (1 + x^{2})} > e^{- π / 2} \\ 2 + e^{- π / 4} & \geq 2 + e^{- \tan^{- 1} (1 + x^{2})} > 2 + e^{- π / 2} \end{aligned}$

To find the derivative of an inverse trig function, set $y = \sin^{- 1} (x)$ , $x = \sin y$ , diff wrt x, use chain rule
- obtain $$
  \begin{align}
  (\sin ^{-1}(x))' & =\frac{1}{\sqrt{ 1-x^{2} }} \
  (\cos ^{-1}(x))' & =-\frac{1}{\sqrt{ 1-x^{2} }} \
  (\tan ^{-1}(x))' & =\frac{1}{1+x^{2}}
  \end